Question

Thin symmetric prism of $\mu = 1.5$. Find ratio of incident angle and minimum deviation.


 

Answer 1

Given:
Refractive index μ = 1.5
Thin symmetric prism

Step 1: Relation at minimum deviation
For prism:
μ = sin[(A + δ)/2] / sin(A/2)

For thin prism approximation:
δ ≈ (μ − 1)A

Step 2: For symmetric condition
Angle of incidence i ≈ A/2 + δ/2

Step 3: Substitute δ
i ≈ A/2 + (μ − 1)A/2
= A/2 (1 + μ − 1)
= (μA)/2

Step 4: Ratio i : δ
\[ \frac{i}{\delta} = \frac{(\mu A)/2}{(\mu - 1)A} = \frac{\mu}{2(\mu - 1)} \]

Step 5: Substitute μ = 1.5
\[ \frac{i}{\delta} = \frac{1.5}{2(0.5)} = \frac{1.5}{1} = \frac{3}{2} \]

Final Answer: 3/2